Question Number 167586 by cortano1 last updated on 20/Mar/22
![λ=∫ (dx/( (√(1+cos x))+(√(1+sin x)))) =?](https://www.tinkutara.com/question/Q167586.png)
Commented by MJS_new last updated on 20/Mar/22
![1^(st) step t=tan (x/2) 2^(nd) step u=t+(√(t^2 +1)) I′ve got no time right now...](https://www.tinkutara.com/question/Q167605.png)
Answered by greogoury55 last updated on 20/Mar/22
![∫(((√(1+cos x))−(√(1+sin x)))/(cos x−sin x)) dx = ∫ ((√(2cos^2 ((x/2))))/(cos x−sin x)) dx−∫ ((√((cos (1/2)x+sin (1/2)x)^2 ))/(cos x−sin x)) dx = ∫ (((√2) cos (x/2))/(cos x−sin x)) dx−∫ ((cos (x/2)+sin (x/2))/(cos x−sin x)) dx](https://www.tinkutara.com/question/Q167589.png)
Answered by MJS_new last updated on 20/Mar/22
![∫(dx/( (√(1+cos x))+(√(1+sin x))))= [t=tan (x/2) → dx=((2dt)/(t^2 +1))] =2∫(dt/((t+1+(√2))(√(t^2 +1))))= [u=t+(√(t^2 +1)) → dt=((√(t^2 +1))/u)du] =4∫(du/(u^2 +2(1+(√2))u−1))= =(√(2−(√2)))ln ((u+1+(√2)−(√(4+2(√2))))/(u+1+(√2)+(√(4+2(√2))))) =...](https://www.tinkutara.com/question/Q167610.png)
Commented by peter frank last updated on 21/Mar/22
![thank you](https://www.tinkutara.com/question/Q167623.png)